Why is it that given $m >n > 0$ and a Lebesgue measurable set of measure $m$, it must have a Lebesgue measurable subset of size $n$?

I had the following idea in mind : Calling our set $A$, since it is Lebesgue measurable, we know $\lambda(A) = \inf\{\sum_{k} l(I_k) : I_k \mbox{ is an interval cover of A}\}$. However, this is a bit of a definition from outside, so we cannot manipulate this definition to get subsets of $A$.

At the same time, $\lambda(A)$ is also the supremum over the measure of all compact subsets of the set. By this logic, we can find a compact set with measure $ > 1 - \epsilon$ for any $\epsilon > 0$, but how about being equal to a certain number? It's not clear to me.

  • $\begingroup$ What about something like defining a function $ F : [0, \infty) \to 2^{A}$ as $F(t) := A \setminus (-t, t)$. Then I belive that it shouldn't be hard to prove that $t \mapsto \lambda(F(t))$ is a continuous function. Moreover $\lambda(F(0)) = m$ and $\lim_{t \rightarrow \infty} \lambda(F(0)) = 0$. Therefore there must be $t_0 >0$ such that $\lambda(F(t)) = n$. $\endgroup$ – Onil90 May 12 '18 at 13:50
  • $\begingroup$ I see. Thanks for the input. $\endgroup$ – Teresa Lisbon May 12 '18 at 13:53

Assume $A\subset \mathbb{R}^d$ and define $f(\lambda)=\mathcal{L}^d(\{x=(x_1,...x_d)\in A \vert ~ x_1<\lambda\})$ for $\lambda\in \mathbb{R}$. Show that $f$ is a continuous function and use the intermediate value theorem.

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  • $\begingroup$ Continuity of $f$ should follow easily from the fact that the strip between $f(\lambda)$ and $f(\mu)$ has volume $|\mu - \lambda|$. Great, thanks for that. $\endgroup$ – Teresa Lisbon May 12 '18 at 13:54
  • $\begingroup$ This would only work for $d=1$, because otherwise the strips have infinite volume. You can instead argue that (denoting $A_\lambda:=A\cap\{x_1<\lambda\}$) for a fixed $\lambda_0$ for any sequence $(\lambda_k)_k$ approaching $\lambda_0$ from below, you have $A_{\lambda_0}=\bigcup_{k\in\mathbb{N}}A_{\lambda_k}$ and thus $f(\lambda_0)=\mathcal{L}^d(A_{\lambda_0})=\lim_{n\rightarrow\infty} \mathcal{L}^d(\bigcup_{k=1}^n A_{\lambda_k})=\lim_n f(\lambda_n)$ and so on. (Which follows directly from the measure axioms.) $\endgroup$ – Jan Bohr May 12 '18 at 14:04
  • $\begingroup$ Thank you for the correction. $\endgroup$ – Teresa Lisbon May 12 '18 at 14:06

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